| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Variable force along axis work-energy |
| Difficulty | Standard +0.8 This M3 question requires integration of a variable force with time, applying impulse-momentum theorem to find limiting speed, then integrating velocity for displacement and calculating kinetic energy changes. While the integrations are straightforward (rational functions), the multi-step reasoning connecting force→velocity→displacement→energy across three parts, plus recognizing the limiting behavior as t→∞, elevates this above routine mechanics problems. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.02c Work by variable force: using integration6.06a Variable force: dv/dt or v*dv/dx methods |
**Part (a)**
B1 for a correct equation of motion with acceleration $= \frac{dv}{dt}$. Can be awarded by implication if work correct at next stage.
$$0.4 \frac{dv}{dt} = \frac{4}{(t+5)^2}$$
M1 for attempting the integration wrt $t$ to obtain an expression for $v$
$$v = \frac{10}{t+5} + c$$
A1 for correct result, constant not needed
M1 dep for using $t = 0, v = 4$ to obtain a value for $c$. Dependent on the previous M mark.
$c = 6$
$$v = 6 + \frac{10}{t+5} \quad t \geq 0$$
or
$$v = 6 - \frac{10}{t+5}$$
A1 cso for a correct concluding statement. Can have $\geq$ or $>$
**Part (b)**
M1 for attempting the integration of their expression for $v$. Limits need not be seen for this mark.
$$s = 6t + 10\ln(t+5) + c$$
A1 ft for correct integration
M1 for substituting the limits 2 and 7
$$42 + 10\ln 12 - 10\ln 7 = 30 + 10\ln\left(\frac{12}{7}\right)$$ or eg $24.6100...$
A1 cao for a correct result, exact or decimal (min 2 sf) $\approx 25$ or better
**Part (c)**
M1 for attempting the difference of KE between the points A and B (either way round). Velocities to be calculated using their expression for $v$. Award for a gain or a loss.
$$\text{KE} = \frac{1}{2}(0.4)\left(6 + \frac{10}{12}\right)^2 - \frac{1}{2}(0.4)\left(6 + \frac{10}{7}\right)^2$$
A1 ft for KE at B - KE at A, with their expression for $v$. Need not be simplified, may be reversed.
A1 cso for $1.1592...$ J Accept 1.2 or better. Must be positive.
---\begin{enumerate}
\item At time $t = 0$, a particle $P$ of mass 0.4 kg is at the origin $O$ moving with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ along the $x$-axis in the positive $x$ direction. At time $t$ seconds, $t \geqslant 0$, the resultant force acting on $P$ has magnitude $\frac { 4 } { ( t + 5 ) ^ { 2 } } \mathrm {~N}$ and is directed away from $O$.\\
(a) Show that the speed of $P$ cannot exceed $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\end{enumerate}
The particle passes through the point $A$ when $t = 2$ and passes through the point $B$ when $t = 7$\\
(b) Find the distance $A B$.\\
(c) Find the gain in kinetic energy of $P$ as it moves from $A$ to $B$.\\
\hfill \mbox{\textit{Edexcel M3 2014 Q4 [12]}}